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Thread: Unity in Subrings, Zero Divisors

  1. #1
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    Unity in Subrings, Zero Divisors

    Hello!

    I have two questions!

    (1)

    if $\displaystyle M$ is a subring of $\displaystyle N$, where $\displaystyle N$ has unity (let's call it $\displaystyle 1_N$), and say $\displaystyle 1_M$ is the unity of $\displaystyle M$. Now, I have shown that it is NOT necessarily true that $\displaystyle 1_M = 1_N$

    Then if $\displaystyle M$ and $\displaystyle N$ are both fields (every element is a unit) then I'm trying to show that it is necessarily true that $\displaystyle 1_M = 1_N$.

    I am thinking you just take any $\displaystyle m \in M$, then $\displaystyle m*m^{-1} = 1_M$, but since $\displaystyle m \in N$ as well, $\displaystyle m*m^{-1} = 1_N$. Is that legitimate reasoning? I'm not confident with my answer for some reason.

    (2) Let $\displaystyle m,n \in R$ for a commutative ring $\displaystyle R$.

    Suppose that $\displaystyle mn$ is a zero divisor in $\displaystyle R$. Prove that either $\displaystyle m$ or $\displaystyle n$ is a zero divisor.

    So, Well, first what I did was use the defn of zero divisor (z.d.)
    If $\displaystyle mn \neq 0$ is a z.d. then $\displaystyle \exists x \in R$ such that $\displaystyle x \neq 0 $ and $\displaystyle mn(x) = 0$

    Then, just by associativity, $\displaystyle m(nx) = 0$ so, well, can we assume that $\displaystyle nx \neq 0$ and $\displaystyle m \neq 0$? Cause we'd be done - $\displaystyle m$ is a z.d. ! By commutativity, just switch $\displaystyle m$ and $\displaystyle n$ and then we're done.

    Any help appreciated! Thanks!!
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  2. #2
    Senior Member roninpro's Avatar
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    Your proof for (1) looks okay. You just need to emphasize that inverses are unique.

    You have a good idea for (2). To clean it up, you should note that neither $\displaystyle m$ nor $\displaystyle n$ should be zero, or else $\displaystyle mn$ is zero, which is not what we want. Then, you are correct in taking nonzero $\displaystyle x$ so that $\displaystyle mnx=0$ and $\displaystyle m(nx)=0$. So if $\displaystyle nx\ne 0$, then you are done, since $\displaystyle m$ will satisfy the zero divisor definition. But what happens if $\displaystyle nx=0$?
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  3. #3
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    I'd just like to point out that (1) is actually true whenever the rings are integral domains.

    In the ring $\displaystyle M$, of course $\displaystyle 1_M\cdot 1_M=1_M$. Also, since $\displaystyle 1_M\in N$ and $\displaystyle 1_N$ is the multiplicative identity of $\displaystyle N$, we get $\displaystyle 1_M\cdot 1_N=1_M$.

    Put these equalities together to get $\displaystyle 1_M\cdot 1_M=1_M\cdot 1_N$. Then, since we are in an integral domain, we can "cancel" the $\displaystyle 1_M$ from both sides to conclude $\displaystyle 1_M=1_N$.
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